I know that a map $\phi:(M,g)\to (N,h)$ between Riemannian manifolds is horizontally weakly conformal at $p\in M$ if and only if the differential $d\phi_p=0$ or $d\phi_p$ maps the horizontal space $\{\ker{d\phi_p}\}^\perp$ conformally onto the tangent space $T_{\phi(p)}N$.
How does one apply the above to the case when $\phi$ is a complex map on $\mathbb{R}^3$? That is, when $M=\mathbb{R}^3$ and $N=\mathbb{C}$. In the literature I have read, they state that such a map is horizontally weakly conformal if and only if $$\sum_{i=1}^3\left(\frac{\partial \phi}{\partial x_i}\right)^2=0, $$ or, more compactly, $\langle\nabla\phi,\nabla\phi\rangle=0$. I am not sure how one sees that this is equivalent to saying that the scalar product between two vector fields $X,Y$ on the horizontal part of $\mathbb{R}^3$ is the same as the scalar product between $d\phi_p(X)$ and $d\phi_p(Y)$ on $\mathbb{C}$.